CH2N2 Lewis structure

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CH2N2 Lewis Structure
CH2N2 Lewis structure

CH2N2 (diazomethane) has one carbon atom, two hydrogen atoms, and two nitrogen atoms.

In the CH2N2 Lewis structure, there is a double bond between the carbon and nitrogen atom. The carbon atom is attached with two hydrogen atoms, and the nitrogen atom is attached with one other nitrogen atom. And on the right nitrogen atom, there are two lone pairs.

Also, there is a negative (-1) charge on the right nitrogen atom, and a positive (+1) charge on the left nitrogen atom.

Steps

Here’s how you can draw the CH2N2 Lewis structure step by step:

#1 Draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Mark charges on the atoms
#4 Minimize charges by converting lone pairs
#5 If central atom doesn’t form octet, convert lone pair and mark charges again

Let’s break down each step in detail.

#1 Draw a rough sketch

  • First, determine the total number of valence electrons
Periodic table

In the periodic table, carbon lies in group 14, hydrogen lies in group 1, and nitrogen lies in group 15.

Hence, carbon has four valence electrons, hydrogen has one valence electron, and nitrogen has five valence electrons.

Since CH2N2 has one carbon atom, two hydrogen atoms, and two nitrogen atoms, so…

Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of two nitrogen atoms = 5 × 2 = 10

And the total valence electrons = 4 + 2 + 10 = 16

  • Second, find the total electron pairs

We have a total of 16 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 16 ÷ 2 = 8

  • Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from carbon and nitrogen. Place the least electronegative atom at the center.

Since carbon is less electronegative than nitrogen, assume that the central atom is carbon.

Therefore, place carbon in the center and hydrogen and nitrogen on either side.

  • And finally, draw the rough sketch
CH2N2 Lewis Structure (Step 1)
Rough sketch of CH2N2 Lewis structure

#2 Mark lone pairs on the atoms

Here, we have a total of 8 electron pairs. And four bonds are already marked. So we have to only mark the remaining four electron pairs as lone pairs on the sketch.

Also remember that both (carbon and nitrogen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell. And hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and nitrogens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for left nitrogen, there are two lone pairs, for right nitrogen, there are three lone pairs, and for carbon, there is zero lone pair because all four electron pairs are over.

Mark the lone pairs on the sketch as follows:

CH2N2 Lewis Structure (Step 2)
Lone pairs marked on CH2N2 Lewis structure

#3 Mark charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For carbon atom, formal charge = 4 – 0 – ½ (6) = +1

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For left nitrogen atom, formal charge = 5 – 2 – ½ (4) = +1

For right nitrogen atom, formal charge = 5 – 6 – ½ (2) = -2

Here, both carbon and nitrogen atoms have charges, so mark them on the sketch as follows:

CH2N2 Lewis Structure (Step 3)
Formal charges marked on CH2N2 Lewis structure

The above structure is not a stable Lewis structure because both carbon and nitrogen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

#4 Minimize charges by converting lone pairs

Convert a lone pair of the right nitrogen atom to make a new N — N bond with the left nitrogen atom as follows:

CH2N2 Lewis Structure (Step 4)
Lone pair of right nitrogen is converted, but central atom doesn’t form octet

In the above structure, you can see that the central atom (carbon) doesn’t form an octet. Hence, the octet rule is not satisfied.

#5 Convert lone pair and mark charges again

Therefore, again convert a lone pair of the left nitrogen atom to make a new C — N bond with the carbon atom. Also, we have to again check whether there are charges on atoms or not.

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For carbon atom, formal charge = 4 – 0 – ½ (8) = 0

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For left nitrogen atom, formal charge = 5 – 0 – ½ (8) = +1

For right nitrogen atom, formal charge = 5 – 4 – ½ (4) = -1

Here, both nitrogen atoms have charges, so mark them on the sketch as follows:

CH2N2 Lewis Structure (Step 5)
Lone pair of left nitrogen is converted, and got the most stable Lewis structure of CH2N2

In the above structure, you can see that the central atom (carbon) forms an octet. The outside atoms (nitrogens) also form an octet, and both hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Now there are still charges on the atoms. But we can not convert a lone pair to a bond because nitrogen can not keep more than 8 electrons in its last shell.

The formal charges on atoms are closer to zero. Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable Lewis structure of CH2N2.

Next: SiBr4 Lewis structure

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Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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